What is the arclength of f(x)=e^(1/x)/x on x in [1,2]?

1 Answer
Jan 9, 2016

Wolframalpha states that the integral you would have to encounter for the determination of the arc length doesn't have a solution in terms of standard mathematical functions.

That just means that you are only expected to evaluate it numerically (or get to the integrand without evaluating it).

A quick derivation on the arc length reveals that it came from a derivatives treatment of the distance formula.

Arc Length:
$D \left(x\right) = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$

$s = D \left(x\right) = \sum \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} / {\left(\Delta x\right)}^{2} \cdot {\left(\Delta x\right)}^{2}}$

$= \sum \sqrt{1 + {\left(\frac{\Delta y}{\Delta x}\right)}^{2}} \left(\Delta x\right)$

$= \textcolor{b l u e}{{\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}}$

This is just a "dynamic", infinitesimally-short-distance formula that accumulates over an interval of constantly increasing $x$. The general strategy is to get common denominators, perhaps complete the square, and get the square root to go away.

So, you can see that the first thing we could do is take the derivative and then square it.

$\frac{d}{\mathrm{dx}} \left[{e}^{\text{1/x}} \cdot \frac{1}{x}\right]$

$= {e}^{\text{1/x"*(-1/x^2) + 1/x * e^"1/x}} \cdot \left(- \frac{1}{x} ^ 2\right)$

$= - \frac{{e}^{\text{1/x")/(x^2) - (e^"1/x}}}{{x}^{3}}$

Okay... yeah. Let's square... this.

${\left(- \frac{{e}^{\text{1/x")/(x^2) - (e^"1/x}}}{{x}^{3}}\right)}^{2}$

$= \left[- \frac{{e}^{\text{1/x")/(x^2) - (e^"1/x")/(x^3)]*[-(e^"1/x")/(x^2) - (e^"1/x}}}{{x}^{3}}\right]$

$= \left[\frac{{e}^{\text{1/x")/(x^2) + (e^"1/x")/(x^3)][(e^"1/x")/(x^2) + (e^"1/x}}}{{x}^{3}}\right]$

$= \frac{{e}^{\text{2/x")/(x^4) + (2e^"2/x")/(x^5) + (e^"2/x}}}{{x}^{6}}$

Now, let's simplify as much as possible. How about common denominators by multiplying until we get ${x}^{6}$ as the denominator?

$= \frac{{x}^{2} {e}^{\text{2/x")/(x^6) + (2xe^"2/x")/(x^6) + (e^"2/x}}}{{x}^{6}}$

$= {e}^{\text{2/x}} \left[\frac{{x}^{2}}{{x}^{6}} + \frac{2 x}{{x}^{6}} + \frac{1}{{x}^{6}}\right]$

Since ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$, we now have:

$= \textcolor{g r e e n}{\frac{{e}^{\text{2/x}} {\left(x + 1\right)}^{2}}{x} ^ 6}$

But Wolframalpha states that this integral doesn't have a solution in terms of standard mathematical functions.

$s = {\int}_{1}^{2} \sqrt{1 + {\left(- \frac{{e}^{\text{1/x")/(x^2) - (e^"1/x}}}{{x}^{3}}\right)}^{2}} \mathrm{dx}$

$= \textcolor{b l u e}{{\int}_{1}^{2} \sqrt{1 + \frac{{e}^{\text{2/x}} {\left(x + 1\right)}^{2}}{x} ^ 6} \mathrm{dx}}$

That just means that you are only expected to evaluate this numerically (or get to the integrand without evaluating it).

$s = {\int}_{1}^{2} \sqrt{\frac{{x}^{6} + {e}^{\text{2/x}} {\left(x + 1\right)}^{2}}{x} ^ 6} \mathrm{dx}$

$\approx \textcolor{b l u e}{2.2103}$